NSWChemistrySyllabus dot point

Inquiry Question 3: It is all about hydrogen ions

Calculate concentration changes on dilution using c1v1 = c2v2 and predict the effect of dilution on pH for strong and weak acid and base solutions

Q: 2022 HSC HSC: A 25.0 mL aliquot of 0.500 mol/L HCl is diluted to a final volume of 250.0 mL with distilled water. Calculate (a) the concentration of the diluted solution and (b) the pH before and after dilution. (c) State, without further calculation, how the pH would have changed if the original acid had been 0.500 mol/L ethanoic acid (Ka = 1.8 x 10^-5).

**(a) Dilution.** Apply $c_1V_1 = c_2V_2$: $$c_2 = \frac{c_1 V_1}{V_2} = \frac{0.500 \times 25.0}{250.0} = 0.0500 \text{ mol/L}$$ **(b) pH before and after.** HCl is a strong acid, so $[H^+] = c$. $$pH_1 = -\log_{10}(0.500) = 0.301$$ $$pH_2 = -\log_{10}(0.0500) = 1.301$$ A 10-fold dilution raises the pH by exactly 1 unit for a strong acid (because $\log_{10}(10) = 1$). **(c) For ethanoic acid.** A 10-fold dilution would raise the pH by **less than 1 unit**. The reason: as the weak acid is diluted, the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts right (Le Chatelier favours the side with more particles), so the percent ionisation increases. $[H^+]$ falls by a factor smaller than 10, so the pH change is smaller than 1. Quantitatively, $[H^+] \approx \sqrt{K_a c}$, so a 10-fold drop in $c$ gives only a $\sqrt{10} \approx 3.2$-fold drop in $[H^+]$, a pH rise of about 0.5. Markers reward (1) correct $c_1V_1 = c_2V_2$, (2) two correct pH values with the right number of sig figs, (3) recognising the smaller pH change for the weak acid with a Le Chatelier or $\sqrt{K_a c}$ justification.

Q: 2017 HSC HSC: A student needs to prepare 500.0 mL of 0.100 mol/L NaOH from a stock 2.00 mol/L NaOH solution. Describe the procedure, including the volume of stock required, the apparatus used, and how the volume is made up accurately.

**Step 1: Volume of stock.** $$V_1 = \frac{c_2 V_2}{c_1} = \frac{0.100 \times 500.0}{2.00} = 25.0 \text{ mL}$$ **Procedure.** 1. Use a 25.0 mL bulb pipette to transfer 25.0 mL of the stock NaOH into a 500.0 mL volumetric flask. 2. Add distilled water to about three-quarters full and swirl to mix. 3. Top up to the calibration mark with distilled water, using a dropper for the final few drops so that the bottom of the meniscus sits on the line at eye level. 4. Stopper and invert the flask several times to mix thoroughly. Markers reward (1) the correct volume from $c_1V_1 = c_2V_2$, (2) named volumetric glassware (bulb pipette and volumetric flask), (3) correct meniscus reading and mixing technique.

A focused answer to the HSC Chemistry Module 6 dot point on dilution. Concentration units (mol/L, percent w/v, ppm), the dilution equation c1v1 = c2v2, how pH changes on dilution for strong and weak acids, and worked HSC past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-18

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to handle the quantitative side of acid-base chemistry: convert between concentration units, perform serial and one-step dilutions, and predict how pH changes when an acid or base is diluted. The mathematics is the dilution identity $c_1 V_1 = c_2 V_2$ ; the chemistry is the difference between strong and weak acids on dilution, which builds on the strong vs weak page and feeds into titration analysis.

The answer

Concentration units

Unit	Symbol	Meaning
Molarity	mol/L (M)	moles of solute per litre of solution
Mass percent	percent w/w	g solute per 100 g solution
Mass/volume percent	percent w/v	g solute per 100 mL solution
Volume percent	percent v/v	mL solute per 100 mL solution
Parts per million	ppm	mg solute per L solution (for dilute aqueous solutions)
Parts per billion	ppb	IMATH_6 g solute per L solution

Molarity is the default HSC unit because it links directly to stoichiometry ( $n = cV$ ). The other units appear in environmental and biological contexts (lead in drinking water, residual chlorine, salinity).

Conversions to remember.

IMATH_8 for dilute aqueous solutions (because 1 L of water has mass 1 kg).
IMATH_9 , where $M_r$ is the molar mass in g/mol.

The dilution equation

When solvent is added to a solution, the moles of solute do not change. Therefore:

n_1 = n_2 \quad \Rightarrow \quad c_1 V_1 = c_2 V_2

Use any consistent volume unit (mL or L) as long as both sides match. The equation works for any solute, not just acids and bases.

Procedure for an accurate dilution.

Calculate the volume of stock needed: $V_1 = c_2 V_2 / c_1$ .
Transfer $V_1$ to a volumetric flask of size $V_2$ using a pipette (for small volumes) or a measuring cylinder.
Add distilled water to roughly three-quarters full, swirl.
Top up to the calibration mark, with a dropper for the last few drops.
Stopper, invert several times to ensure homogeneity.

pH on dilution: strong acid or base

For a strong acid, $[H^+] = c$ (fully ionised). On dilution, $[H^+]$ drops in direct proportion to $c$ . A 10-fold dilution raises pH by 1.0 unit; a 100-fold dilution raises pH by 2.0 units.

For a strong base, $[OH^-] = c$ on the same logic, and pH falls by 1 unit per 10-fold dilution (because pOH rises by 1).

Limit: as you dilute past about $10^{-6}$ mol/L, the auto-ionisation of water becomes significant and pH approaches 7 from below (for an acid) or above (for a base), never crossing 7. A common HSC error is to claim that $10^{-9}$ mol/L $HCl$ has pH 9; in fact it has pH just under 7 because the dominant source of $H^+$ is now water itself.

pH on dilution: weak acid or base

For a weak acid, $[H^+] \approx \sqrt{K_a c}$ (from the ICE shortcut, valid when ionisation is small).

A 10-fold dilution changes $c$ by 10, so $[H^+]$ changes by $\sqrt{10} \approx 3.16$ , and pH rises by $\log_{10}(\sqrt{10}) = 0.5$ . Weak acids resist pH change on dilution more than strong acids do.

Mechanistically (Le Chatelier): the ionisation equilibrium has more moles of dissolved particles on the right side, so dilution favours further ionisation. The percent ionisation rises as concentration falls, partially offsetting the dilution.

Summary rule. A 10-fold dilution raises pH by:

1.0 unit for a strong acid (until water dominates).
~0.5 unit for a weak acid.
~0 for a buffer (until exhausted).

Serial dilution

If you need a very dilute solution, a single one-step dilution can be inaccurate (pipetting a very small volume). Serial dilution does it in stages: each step is a 10- or 100-fold dilution. To go from 1.00 mol/L to $1.00 \times 10^{-4}$ mol/L, do four successive 10-fold dilutions.

Worked example

Calculate the pH of (a) $0.0100$ mol/L $HCl$ and (b) $0.0100$ mol/L $CH_3COOH$ ( $K_a = 1.8 \times 10^{-5}$ ), then state the pH that each would have after a 100-fold dilution.

(a) HCl, initial. $[H^+] = 0.0100$ mol/L, $pH = 2.00$ .

(a) HCl after 100-fold dilution. $c = 1.00 \times 10^{-4}$ mol/L, $pH = 4.00$ .

(b) Ethanoic acid, initial.

[H^+] \approx \sqrt{(1.8 \times 10^{-5})(0.0100)} = \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4} \text{ mol/L}

pH = -\log_{10}(4.24 \times 10^{-4}) = 3.37

(b) Ethanoic acid after 100-fold dilution. $c = 1.00 \times 10^{-4}$ mol/L:

[H^+] \approx \sqrt{(1.8 \times 10^{-5})(1.00 \times 10^{-4})} = \sqrt{1.8 \times 10^{-9}} = 4.24 \times 10^{-5} \text{ mol/L}

pH = -\log_{10}(4.24 \times 10^{-5}) = 4.37

The strong acid rose by 2.0 pH units (as predicted: 100-fold dilution, $\log_{10}(100) = 2$ ). The weak acid rose by 1.0 pH unit (as predicted: $\log_{10}(\sqrt{100}) = 1$ ). At this point the weak acid pH (4.37) is higher than the strong acid pH (4.00) because the strong acid is still fully ionised and the weak acid is not, even though both share the same nominal concentration.

Common traps

Forgetting consistent units. In $c_1 V_1 = c_2 V_2$ , both volumes must be in the same unit. Mixing mL and L drops a factor of 1000.

Applying $[H^+] = c$ to a weak acid. That formula is for strong acids only. Use $\sqrt{K_a c}$ for weak acids.

Extrapolating strong-acid pH past auto-ionisation. $10^{-9}$ mol/L $HCl$ does not have pH 9. Below about $10^{-6}$ mol/L, water's auto-ionisation contributes; a full charge balance gives pH approaching 7 from the acidic side.

Calling dilute strong acid "weak". Dilution lowers concentration; it does not change the strength (degree of ionisation). A dilute strong acid is still a strong acid (still fully ionised).

Confusing percent w/v with mol/L. A 5 percent w/v $NaOH$ solution is 50 g/L, which converts to $50/40 = 1.25$ mol/L. Always carry the conversion through molar mass.

Pipetting from a volumetric flask without rinsing. When making a standard, the pipette should be rinsed with the stock first (not water), and the volumetric flask should be rinsed with distilled water (not stock).

In one sentence

Moles of solute are conserved on dilution, so $c_1 V_1 = c_2 V_2$ gives the new concentration; a 10-fold dilution raises strong-acid pH by exactly 1 unit (because $[H^+] = c$ ) but raises weak-acid pH by only about 0.5 units (because $[H^+] \approx \sqrt{K_a c}$ and the equilibrium shifts right on dilution), and very dilute strong acid pH never exceeds 7 because water's auto-ionisation eventually dominates.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC4 marksA 25.0 mL aliquot of 0.500 mol/L HCl is diluted to a final volume of 250.0 mL with distilled water. Calculate (a) the concentration of the diluted solution and (b) the pH before and after dilution. (c) State, without further calculation, how the pH would have changed if the original acid had been 0.500 mol/L ethanoic acid (Ka = 1.8 x 10^-5).

Show worked answer →

(a) Dilution. Apply $c_1V_1 = c_2V_2$ :

c_2 = \frac{c_1 V_1}{V_2} = \frac{0.500 \times 25.0}{250.0} = 0.0500 \text{ mol/L}

(b) pH before and after. HCl is a strong acid, so $[H^+] = c$ .

pH_1 = -\log_{10}(0.500) = 0.301

pH_2 = -\log_{10}(0.0500) = 1.301

A 10-fold dilution raises the pH by exactly 1 unit for a strong acid (because $\log_{10}(10) = 1$ ).

(c) For ethanoic acid. A 10-fold dilution would raise the pH by less than 1 unit. The reason: as the weak acid is diluted, the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts right (Le Chatelier favours the side with more particles), so the percent ionisation increases. $[H^+]$ falls by a factor smaller than 10, so the pH change is smaller than 1.

Quantitatively, $[H^+] \approx \sqrt{K_a c}$ , so a 10-fold drop in $c$ gives only a $\sqrt{10} \approx 3.2$ -fold drop in $[H^+]$ , a pH rise of about 0.5.

Markers reward (1) correct $c_1V_1 = c_2V_2$ , (2) two correct pH values with the right number of sig figs, (3) recognising the smaller pH change for the weak acid with a Le Chatelier or $\sqrt{K_a c}$ justification.

2017 HSC3 marksA student needs to prepare 500.0 mL of 0.100 mol/L NaOH from a stock 2.00 mol/L NaOH solution. Describe the procedure, including the volume of stock required, the apparatus used, and how the volume is made up accurately.